Problem 34 - Algebra 2

Question

Find the exact values of \(\sin 2 \theta, \cos 2 \theta, \sin \frac{\theta}{2},\) and \(\cos \frac{\theta}{2}\) for each of the following. $$ \cos \theta=\frac{2}{3} ; 0^{\circ}<\theta<90^{\circ} $$

Step-by-Step Solution

Answer

\(\sin 2\theta = \frac{4\sqrt{5}}{9}\), \(\cos 2\theta = -\frac{1}{9}\), \(\sin \frac{\theta}{2} = \frac{\sqrt{6}}{6}\), \(\cos \frac{\theta}{2} = \frac{\sqrt{30}}{6}\).

Step 1: Determine Missing Sine Value

Since we are given \( \cos \theta = \frac{2}{3} \) and need to find \( \sin \theta \), we can use the Pythagorean identity: \\[ \sin^2 \theta + \cos^2 \theta = 1 \]\Plug \( \cos \theta = \frac{2}{3} \) into the equation: \\[ \sin^2 \theta + \left( \frac{2}{3} \right)^2 = 1 \]\\[ \sin^2 \theta + \frac{4}{9} = 1 \]\Subtract \( \frac{4}{9} \) from both sides: \\[ \sin^2 \theta = 1 - \frac{4}{9} \]\\[ \sin^2 \theta = \frac{5}{9} \]\Take the square root: \\[ \sin \theta = \frac{\sqrt{5}}{3} \] (since \( 0^\circ < \theta < 90^\circ \), \(\sin \theta\) is positive)

Step 2: Find \(\sin 2\theta\)

Using the double-angle identity for sine: \\[ \sin 2\theta = 2 \sin \theta \cos \theta \]\Substitute \( \sin \theta = \frac{\sqrt{5}}{3} \) and \( \cos \theta = \frac{2}{3} \): \\[ \sin 2\theta = 2 \left( \frac{\sqrt{5}}{3} \right) \left( \frac{2}{3} \right) \]\\[ \sin 2\theta = 2 \times \frac{2\sqrt{5}}{9} = \frac{4\sqrt{5}}{9} \]

Step 3: Find \(\cos 2\theta\)

Use the double-angle identity for cosine: \\[ \cos 2\theta = 2 \cos^2 \theta - 1 \]\Substitute \( \cos \theta = \frac{2}{3} \): \\[ \cos 2\theta = 2 \left( \frac{2}{3} \right)^2 - 1 \]\\[ \cos 2\theta = 2 \times \frac{4}{9} - 1 \]\\[ \cos 2\theta = \frac{8}{9} - 1 = \frac{8}{9} - \frac{9}{9} \]\\[ \cos 2\theta = -\frac{1}{9} \]

Step 4: Find \(\sin \frac{\theta}{2}\)

Using the half-angle formula for sine: \\[ \sin \frac{\theta}{2} = \sqrt{\frac{1 - \cos \theta}{2}} \]\Substitute \( \cos \theta = \frac{2}{3} \): \\[ \sin \frac{\theta}{2} = \sqrt{\frac{1 - \frac{2}{3}}{2}} \]\\[ \sin \frac{\theta}{2} = \sqrt{\frac{\frac{1}{3}}{2}} \]\\[ \sin \frac{\theta}{2} = \sqrt{\frac{1}{6}} = \frac{\sqrt{6}}{6} \]

Step 5: Find \(\cos \frac{\theta}{2}\)

Using the half-angle formula for cosine: \\[ \cos \frac{\theta}{2} = \sqrt{\frac{1 + \cos \theta}{2}} \]\Substitute \( \cos \theta = \frac{2}{3} \): \\[ \cos \frac{\theta}{2} = \sqrt{\frac{1 + \frac{2}{3}}{2}} \]\\[ \cos \frac{\theta}{2} = \sqrt{\frac{\frac{5}{3}}{2}} \]\\[ \cos \frac{\theta}{2} = \sqrt{\frac{5}{6}} = \frac{\sqrt{30}}{6} \]

Key Concepts

Cosine Double-Angle Identity

The cosine double-angle identity is a useful trigonometric formula that allows you to calculate the cosine of double an angle, expressed as \( \cos 2\theta \). This identity states: \[ \cos 2\theta = 2 \cos^2 \theta - 1 \] It is derived from the cosine addition formula \( \cos(a + b) = \cos a \cos b - \sin a \sin b \), by setting \( a = b = \theta \). This double-angle identity is beneficial in many trigonometric problems as it lets you simplify expressions involving angles. For the exercise provided, where \( \cos \theta = \frac{2}{3} \), the double-angle identity helps find the value of \( \cos 2\theta \). You plug in the value of \( \cos \theta \) into the identity, and compute:

• Calculate \( \cos^2 \theta = \left(\frac{2}{3}\right)^2 = \frac{4}{9} \).
• Substitute into the formula: \( 2 \times \frac{4}{9} - 1 = \frac{8}{9} - \frac{9}{9} = -\frac{1}{9} \).

Thus, using this identity reveals that \( \cos 2\theta = -\frac{1}{9} \).

Sine Double-Angle Identity

The sine double-angle identity helps to calculate the sine of double an angle, expressed as \( \sin 2\theta \). It is formulated as: \[ \sin 2\theta = 2 \sin \theta \cos \theta \] This identity is derived from the sine addition formula \( \sin(a + b) = \sin a \cos b + \cos a \sin b \), by choosing \( a = b = \theta \). It is particularly useful for simplifying complex trigonometric expressions and solving equations. In the given exercise, where \( \cos \theta = \frac{2}{3} \) and using the Pythagorean identity to find \( \sin \theta = \frac{\sqrt{5}}{3} \), you can use this identity to determine \( \sin 2\theta \):

• Compute \( \sin 2\theta = 2 \times \frac{\sqrt{5}}{3} \times \frac{2}{3} \).
• This results in \( \sin 2\theta = 2 \times \frac{2\sqrt{5}}{9} = \frac{4\sqrt{5}}{9} \).

The result \( \sin 2\theta = \frac{4\sqrt{5}}{9} \) showcases how effective the use of trigonometric identities can be in solving problems for exact values.

Sine Half-Angle Identity

The sine half-angle identity is used for finding the sine of half an angle, expressed as \( \sin \frac{\theta}{2} \). The identity is given by: \[ \sin \frac{\theta}{2} = \sqrt{\frac{1 - \cos \theta}{2}} \] This formula is very helpful in scenarios where precise values are needed for angles that are halved, such as dividing an angle into two equal parts. From the exercise problem, with \( \cos \theta = \frac{2}{3} \), you can determine \( \sin \frac{\theta}{2} \) as follows:

• Substitute into the formula: \( \sin \frac{\theta}{2} = \sqrt{\frac{1 - \frac{2}{3}}{2}} \).
• Simplify to get \( \sin \frac{\theta}{2} = \sqrt{\frac{1}{6}} = \frac{\sqrt{6}}{6} \).

The identity's ease of use can simplify complicated problems, providing exact values for angles.

Cosine Half-Angle Identity

The cosine half-angle identity is used to calculate the cosine of half an angle, written as \( \cos \frac{\theta}{2} \). The formula for this identity is: \[ \cos \frac{\theta}{2} = \sqrt{\frac{1 + \cos \theta}{2}} \] This identity is incredibly useful for determining the exact values of angles that are halved, especially when working within the constraints of trigonometric problems. Examining the exercise, with \( \cos \theta = \frac{2}{3} \), you can use this identity to find \( \cos \frac{\theta}{2} \):

• Substitute into the formula: \( \cos \frac{\theta}{2} = \sqrt{\frac{1 + \frac{2}{3}}{2}} \).
• Simplify to yield \( \cos \frac{\theta}{2} = \sqrt{\frac{5}{6}} = \frac{\sqrt{30}}{6} \).

Utilizing this identity facilitates solving trigonometry problems by providing manageable expressions and exact solutions.